Maximum ProfitSay you have an array, A, for which the ith element is the price of a given stock on day i.Design an algorithm to find the maximum profit. You may complete at most 2 transactions.Return the maximum possible profit.Note: You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).Problem Constraints1 <= length(A) <= 7e51 <= A[i] <= 1e7Sample input57 2 4 8 7Sample output6ExplanationInput: 57 2 4 8 7The first line contains the number of elements in the array.And, the second line contains the element of array.Output:6

The problem is asking to find the maximum profit we can get by performing at most two transactions. A transaction is buying and selling the stock. We can't perform multiple transactions at the same time, meaning we can't buy before selling the stock.

Here is a step-by-step solution:

1. Initialize two arrays, profit1 and profit2, of the same length as the input array. These arrays will store the maximum profit that can be made with at most one transaction and at most two transactions, respectively.

2. Set profit1[0] to 0 because we can't make any profit on the first day. For profit1[i] where i > 0, it is the maximum of profit1[i-1] (the maximum profit made until the previous day) and A[i] - min_price, where min_price is the minimum price from day 0 to day i. This represents the maximum profit we can make by selling the stock on day i.

3. Set profit2[0] and profit2[1] to 0 because we can't make any profit on the first two days with two transactions. For profit2[i] where i > 1, it is the maximum of profit2[i-1] (the maximum profit made until the previous day) and A[i] - min_price_with_profit, where min_price_with_profit is the minimum value of A[j] - profit1[j] for j from 0 to i-1. This represents the maximum profit we can make by completing the second transaction on day i.

4. The maximum profit that can be made with at most two transactions is profit2[n-1], where n is the length of the input array.

Here is a Python solution:

def maxProfit(A):
    if not A:
        return 0

    profit1 = [0] * len(A)
    profit2 = [0] * len(A)

    min_price = A[0]
    for i in range(1, len(A)):
        min_price = min(min_price, A[i])
        profit1[i] = max(profit1[i-1], A[i] - min_price)

    min_price_with_profit = A[1] - profit1[0]
    for i in range(2, len(A)):
        min_price_with_profit = min(min_price_with_profit, A[i] - profit1[i-1])
        profit2[i] = max(profit2[i-1], A[i] - min_price_with_profit)

    return profit2[-1]

For the sample input 57 2 4 8 7, the output is 6, which is the maximum profit we can make with at most two transactions.

The problem is asking to find the maximum profit we can get by buying and selling the stock at most twice. We can't buy and sell the stock at the same time, we need to sell the stock before we buy again.

Here is a step by step solution:

1. Initialize four variables, firstBuy, firstSell, secondBuy, and secondSell. Set firstBuy and secondBuy to the maximum integer value and firstSell and secondSell to 0.

2. Loop through each price in the price array. For each price, we calculate firstBuy as the minimum of firstBuy and the current price, firstSell as the maximum of firstSell and the current price minus firstBuy, secondBuy as the minimum of secondBuy and the current price minus firstSell, and secondSell as the maximum of secondSell and the current price minus secondBuy.

3. After the loop, secondSell will be the maximum profit we can get by buying and selling the stock at most twice.

Here is the Python code for the above steps:

def maxProfit(prices):
    firstBuy = secondBuy = float('inf')
    firstSell = secondSell = 0
    for price in prices:
        firstBuy = min(firstBuy, price)
        firstSell = max(firstSell, price - firstBuy)
        secondBuy = min(secondBuy, price - firstSell)
        secondSell = max(secondSell, price - secondBuy)
    return secondSell

For the sample input 57 2 4 8 7, the maximum profit is 6. We can buy the stock at price 2 and sell it at price 8, then buy it again at price 7 and sell it at price 7. The total profit is 6 + 0 = 6.